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So now you know that every point in the parametric curve lies on the unit circle, since every point satisfies the equation $x^2+y^2=1$.

Some questions to ask yourself:

Where do you start? That is, when $\theta=0$, where on the unit circle are you?

As $\theta$ increases from $0$ towards $\pi$, what happens to the point $(x,y) = (\sin\theta,\cos\theta)$?

Once $\theta$ reaches $\pi$, where are you?

Now you know which parts of the unit circle are given by the parametric curve.

Added. Like many students early on, you seem to want to plot a few points and then interpolate; this is not a good idea, because it relies on you just happening to hit the correct points to get an accurate picture of what is going on. To give you an example, if you were trying to plot a graph for the function $y = \sin(\pi x)$, and tried a few points, say, $x=0$, $x=1$, $x=2$, $x=3$, etc., you might think that your function is the constant function $0$, because your selection of points happens to miss all the important stuff that is going on with $y$.

You don't want to do that. Instead, you want to think about what those functions are doing.

One of the ways in which I find most fruitful to think about parametric equations is to think of the parameter as giving time, and the equations as describing the movement of a point; imagine an animation with a glowing point moving along the plane, and leaving a "trail" of light behind it. That trail is the parametric curve, the point is the position at the "present $t$". You want to think about what that point is doing as your parameter ranges from its initial value to its final value (that is, as the animation goes from beginning to end).

So, start with $\theta=0$, the first frame of your animation. Your glowing point will be at $x(0) = \sin(0) = 0$, and $y(0)=\cos(0) = 1$. So you start at the point $(0,1)$.

Now, press the PLAY button. What happens as $\theta$ starts advancing from $0$ towards $\pi$? The $x$ coordinate will follow the graph of $y(\theta)=\sin(\theta)$, so first it will rise from $0$ to $1$ (at $\theta=\frac{\pi}{2}$), and then drop again from $1$ to $0$ (at $\theta=\pi$). It will do so without jumps or breaks. So if you were looking only at the "shadow" of our glowing point on the $x$-axis, it starts at $x=0$, then moves towards the right in a smooth way (no jerks, no jumps, no skips) until it hits $1$ at the midpoint of the movie, and then moves back towards $0$ until it returns to $0$ at the end of the movie.

What about $y$? It starts at $1$; it will behave like the graph of $\cos\theta$. As you press PLAY, it will start at $1$, and then drop towards $0$, again in a smooth way, without jumps, jerks, or skips, until it hits $0$ halfway through the movie (at $\theta=\frac{\pi}{2}$). Then, it will keep going in the same direction, from $0$ down to $-1$, and reach $-1$ at the end of the "movie" (when $\theta=\pi$). So if you look at the "shadow" of the glowing point on the $y$-axis, it will start at $1$, then drop down to $0$, and keep going down to $-1$, all in a generally smooth manner, with no jumps, jerks, hesitations, backtracks, etc.

Now put those two motions together: you start at $(0,1)$, the top of the unit circle. Then as $\theta$ moves from $0$ to $\pi/2$, the glowing point starts moving both right and down, always along the unit circle, until $\theta=\pi/2$, when it is at point <fill in the blank>. Press PAUSE on the video, and take stock. Where are we? What part of the unit circle have we traced? How many times? Any backtrack? Any long period of time spent without moving? Was the motion generally "fluid"?

Okay, ready to continue? Press PLAY again, and our $\theta$ starts increasing from $\pi/2$ towards $\pi$. That glowing point representing $(x(\theta),y(\theta))$ moves, but now down and to the left, always along the unit circle, until finally when $\theta=\pi$, the "end of the movie", it reaches its final destination at point <fill in the other blank>.

All along, the movement was without jumps, hesitations, or backtracks, because the functions $x=\sin\theta$ and $y=\cos\theta$ have those motions: no jumps, no skips, no jerks, no hesitations, no backtracks, just a smooth motion (imagine your hand drawing their graphs). That glowing point has now traced part of the unit circle, exactly once, without backtracking. Which part?

oh ok. so essentially, you can just plot the initial point, the ending point to get the shape of the curve, and a point in between to determine the direction of the curve?
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KrystenMar 8 '11 at 3:18

@Krysten: No. Think about what that point is doing as $\theta$ ranges from $0$ to $\pi$; plotting just the initial and terminal point might not give you the entire picture (for example, think about $x = \sin t^2$, $y=\cos t^2$, with $-\sqrt{\pi}\leq t\leq \sqrt{\pi}$; if you just plot initial, terminal, and a point in between, say, $t=0$, you'll miss the picture of what is going on). Imagine that $\theta$ describes time, and the parametric equations give you the point of a pencil moving on the plane. Imagine the pencil moving over time, as $\theta$ goes from $0$ to $\pi$; what does it trace?
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Arturo MagidinMar 8 '11 at 3:21

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@Krysten: Again, no. A table is not going to tell you what is really going on, because you cannot possibly plot enough points to make sure that there isn't any "weirdness" going on "in between" the points you are plotting. There are just too many points. What you want to do is think about the functions. What happens to $\sin \theta$ as $\theta$ ranges from $0$ to $\pi$? It goes up from $0$ through $1$, without skipping or jumping, then after it reaches $1$ at $\pi/2$, it goes back down to $0$, again without skipping or jumping. What does that tell you about the $x$ coordinate? Etc.
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Arturo MagidinMar 8 '11 at 3:32

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@Krysten: $\sin\theta$ is never greater than $1$, so I don't know what it is you think you are seeing now... I'm just going to add to the question, I think you are not seeing the forest for the trees.
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Arturo MagidinMar 8 '11 at 3:41

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@Krysten: Once you get some practice thinking about it this way, it will be pretty fast; in fact, probably faster than plugging in a bunch of points when the functions may be somewhat difficult to compute quantitatively, but easy to understand qualitatively (as $\sin$ and $\cos$ are; we have a very good feel for what they look like, but we have a hard time computing them very well for "most" values).
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Arturo MagidinMar 8 '11 at 4:17